# Prove that for an ideal-gas reaction, (dlnK_C^@)/(dT) = (DeltaU^@)/(RT^2)?

## I found this question particularly difficult, but I figured it out and wanted to share.

Oct 9, 2016

For an ideal gas reaction, we begin with the definition of ${K}_{C}^{\circ}$

${K}_{C}^{\circ} = {K}_{P}^{\circ} {\left(\frac{{P}^{\circ}}{R T {c}^{\circ}}\right)}^{\Delta n}$

and differentiate $\ln {K}_{C}^{\circ}$ with respect to $T$. We would obtain, noting the usage of the product rule on ${K}_{P}^{\circ} = {e}^{- \Delta {G}^{\circ} / \left(R T\right)}$ (since $\Delta {G}^{\circ} = \Delta {G}^{\circ} \left(T\right)$):

$\frac{\mathrm{dl} n {K}_{C}^{\circ}}{\mathrm{dT}}$

$= \frac{d}{\mathrm{dT}} \left[\ln \left\{{e}^{- \frac{\Delta {G}^{\circ}}{R T}} {\left(\frac{{P}^{\circ}}{R T {c}^{\circ}}\right)}^{\Delta n}\right\}\right]$

$= \frac{d}{\mathrm{dT}} \left[- \frac{\Delta {G}^{\circ}}{R T} + \Delta n \ln \left(\frac{{P}^{\circ}}{R T {c}^{\circ}}\right)\right]$

$= \frac{d}{\mathrm{dT}} \left[- \frac{\Delta {G}^{\circ}}{R T}\right] + \Delta n \left(\frac{\cancel{R T {c}^{\circ}}}{\cancel{{P}^{\circ}}}\right) \cdot - \frac{\cancel{{P}^{\circ}}}{\cancel{R {c}^{\circ}} {T}^{\cancel{2}}}$

$= \stackrel{\text{Product Rule}}{\overbrace{\frac{\Delta {G}^{\circ}}{R {T}^{2}} - \frac{1}{R T} \frac{\mathrm{dD} e < a {G}^{\circ}}{\mathrm{dT}}}} - \frac{\Delta n}{T}$

By definition, since $\Delta {G}^{\circ}$ is defined for a fixed $\text{1 bar}$ pressure, we can refer to the Maxwell relation

${\mathrm{dG}}^{\circ} = - {S}^{\circ} \mathrm{dT} + V {\mathrm{dP}}^{\circ}$

or

$\mathrm{dD} e < a {G}^{\circ} = - \Delta {S}^{\circ} \mathrm{dT} + \Delta V {\mathrm{dP}}^{\circ}$

and acquire

${\left(\frac{\partial \Delta {G}^{\circ}}{\partial T}\right)}_{{P}^{\circ}} = \frac{\mathrm{dD} e < a {G}^{\circ}}{\mathrm{dT}} = - \Delta {S}^{\circ}$

Then we proceed to acquire the result by noting that $\Delta {G}^{\circ} = \Delta {H}^{\circ} - T \Delta {S}^{\circ}$, and that for an ideal gas, $\Delta {H}^{\circ} = \Delta {U}^{\circ} + \Delta \left(P V\right) = \Delta {U}^{\circ} + \Delta n R T$:

$\implies \frac{\Delta {G}^{\circ}}{R {T}^{2}} + \frac{\Delta {S}^{\circ}}{R T} - \frac{\Delta n}{T} = \frac{\Delta {G}^{\circ}}{R {T}^{2}} + \frac{T \Delta {S}^{\circ}}{R {T}^{2}} - \frac{\Delta n R T}{R {T}^{2}}$

$= \frac{\Delta {G}^{\circ} + T \Delta {S}^{\circ} - \Delta n R T}{R {T}^{2}} = \frac{\Delta {H}^{\circ} - \cancel{T \Delta {S}^{\circ} + T \Delta {S}^{\circ}} - \Delta n R T}{R {T}^{2}}$

$= \frac{\Delta {U}^{\circ} + \Delta \left(P V\right) - \Delta n R T}{R {T}^{2}} = \frac{\Delta {U}^{\circ} + \cancel{\Delta n R T} - \cancel{\Delta n R T}}{R {T}^{2}}$

$= \textcolor{b l u e}{\frac{\Delta {U}^{\circ}}{R {T}^{2}}}$